What is the solution for y^{''}+9y=csc^2(3t) ?

The solution for y^{''}+9y=csc^2(3t) is y=c_{1}cos(3t)+c_{2}sin(3t)+(-ln(tan((3t)/2))cos(3t)-1)/(9)

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y^{''}+9y=csc^2(3t)

y^{^{''}} + 9 y = csc^{2} (3 t)

y = c_{1} cos (3 t) + c_{2} sin (3 t) + \frac{- ln ( tan ( \frac{3 t}{2} ) ) cos ( 3 t ) - 1}{9}

Solution steps

y^{''} (t) + 9 y = csc^{2} (3 t)

Solve linear ODE:

y = c_{1} cos (3 t) + c_{2} sin (3 t) + \frac{- ln ( tan ( \frac{3 t}{2} ) ) cos ( 3 t ) - 1}{9}

y = c_{1} cos (3 t) + c_{2} sin (3 t) + \frac{- ln ( tan ( \frac{3 t}{2} ) ) cos ( 3 t ) - 1}{9}

y^{^{'}} = 2 y - t^{2}

\int_{0}^{2} sin (x^{2}) d x

\frac{\partial}{\partial x} (- x y z^{2} + 16)

y = x - x^{3} (1)

\frac{5}{s ^{2} + 4}

What is the solution for y^{''}+9y=csc^2(3t) ?
The solution for y^{''}+9y=csc^2(3t) is y=c_{1}cos(3t)+c_{2}sin(3t)+(-ln(tan((3t)/2))cos(3t)-1)/(9)